Find the area of the region enclosed by the inner

Priscilla Johnston 2021-12-20 Answered
Find the area of the region enclosed by the inner loop of the curve.
r=2+4sinθ
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

veiga34
Answered 2021-12-21 Author has 32 answers
Find the interval in which the inner loop exists.
For this we take r=0
2+4sinθ=0
sinθ=12
θ=7π6,11π6
So inner loop lies between the lines θ=7π6 and θ=11π6
Since curve is symmetric about the vertical polar axis. So we canf ind the area of the loop by making double the area between θ=7π6 and θ=3π2
Then Area of the loop is A=27π63π212r2dθ
Therefore
A=7π63π2(2+4sinθ)2dθ
A=47π63π2(1+2sinθ)2dθ
=47π63π2(1+2(1cos2θ)+4sinθ)dθ
=47π63π2(32cos2θ+4sinθ)dθ
=4[3θsin2θ+4(cosθ]7π63π2
=4[9π27π2323]
A=4π63
Not exactly what you’re looking for?
Ask My Question
Archie Jones
Answered 2021-12-22 Author has 34 answers
The inner loop of the curve is bounded by the 2 points where r=0. Plug that in and you get sin? = -1/2, so ? = 210 and 330 degrees, or 7?/6 and 11?/6. So the integral needed is
(2+4sin(θ))dθ
Apply the Sum Rule:
=2dθ+4sin(θ)dthη BSJ 2dθ=2θ
4sin(θ)dθ=4cos(θ)
2θ4cos(θ)
=2θ4cos(θ)+C
Compute the boundaries: 7π611π6(2+4sin(θ))dθ=11π323(23+7π3)
=11π323(23+7π3)
Simplify:
=4π343
Not exactly what you’re looking for?
Ask My Question
RizerMix
Answered 2021-12-29 Author has 438 answers

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more